Past Papers’ Solutions  Edexcel  AS & A level  Mathematics  Core Mathematics 1 (C16663/01)  Year 2012  June  Q#9
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Question
The line L_{1} has equation 4y + 3 = 2x.
The point A (p, 4) lies on L_{1}.
a. Find the value of the constant p.
The line L_{2} passes through the point C (2, 4) and is perpendicular to L_{1}.
b. Find an equation for L_{2} giving your answer in the form ax + by + c = 0, where a, b and c are integers.
The line L_{1} and the line L_{2} intersect at the point D.
c. Find the coordinates of the point D.
d. Show that the length of CD is .
A point B lies on L_{1 }and the length of .
The point E lies on L_{2 }such that the length of the line CDE = 3 times the length of CD.
e. Find the area of the quadrilateral ACBE.
Solution
a.
We are required to find xcoordinate of the point A (p, 4).
If a point lies on the curve (or the line), the coordinates of that point satisfy the equation of the curve (or the line).
We are given that point A lies on the line L_{1 }which has equation;
4y + 3 = 2x
We substitute coordinates of point A in the equation of the line L_{1};
4(4) + 3 = 2(p)
16 + 3 = 2p
2p =19
b.
We are required to find equation of line L_{2}.
To find the equation of the line either we need coordinates of the two points on the line (TwoPoint form of Equation of Line) or coordinates of one point on the line and slope of the line (PointSlope form of Equation of Line).
We have coordinates of a point C(2,4) on line L_{2} but not slope of the line L_{2}.
Next we need slope of the line L_{2 }to write its equation.
We are given that the line L_{2} is perpendicular to L_{1}.
If two lines are perpendicular (normal) to each other, then product of their slopes and is;
Therefore, if we have slope of line L_{1}, we can find slope of the line L_{2}.
We do not have slope of both lines L_{1 }and L_{2}. However, we are given equation of line L_{1} as;
SlopeIntercept form of the equation of the line;
Where is the slope of the line.
Therefore, we rearrange the given equation of the line L_{1} to write it in slopeintercept form;
Therefore;
Hence;
With coordinates of a point on the line L_{2 }as C(2,4) and its slope at hand, we can write
equation of the line L_{2}.
PointSlope form of the equation of the line is;
c.
We are given that the line L_{1} and the line L_{2} intersect at the point D and we are required to find the coordinates of point D.
If two lines (or a line and a curve) intersect each other at a point then that point lies on both lines i.e. coordinates of that point have same values on both lines (or on the line and the curve). Therefore, we can equate coordinates of both lines i.e. equate equations of both the lines (or the line and the curve).
Equation of the line L_{1} is;
Equation of the line L_{2} is;
From equation of the line L_{1}, we can substitute expression for in equation of the line L_{2};
Single value of y indicates that there is only one intersection point.
With ycoordinate of point of intersection of two lines (or line and the curve) at hand, we can find the xcoordinate of the point of intersection of two lines (or line and the curve) by substituting value of ycoordinate of the point of intersection in any of the two equations.
We choose;
Hence coordinates of point D are .
d.
We are required to find the length of CD.
Expression for the distance between two given points and is:
Therefore;
We have coordinates of and . Therefore;
e.
We are given that if point B lies on line L_{1};
We are also given that the point E lies on L_{2} such that the length of the line CDE = 3 times the length of CD.
We are required to find the area of quadrilateral ACBE.
It is becomes much convenient if we sketch the quadrilateral ACBE as shown below.
First of all we draw the line L_{1} passing through point A(9.5,4) and a fictitious point B on the same line (because we donot have coordinates of point B).
Then we draw the line L_{2} which is perpendicular to the line L_{1 }at point D (taken arbitrarily between points A and B) where the two lines intersect as given. Here it is quite possible to draw the line L_{2 }beyond either point A or B outside the segment AB but then it would not make a quadrilateral ACBE.
Next we sketch points C and E on the line L_{2}. Then join points A,B,C and E to make quadrilateral ACBE.
This is shown the figure below.
We are required to find the area of this quadrilateral ACBE.
If we join points A and B, the quadrilateral ACBE is divided into two triangle ABC and ABE.
Therefore;
Expression for the area of the triangle is;
Since CE (the Line L_{2}) and AB (the line L_{1}) are perpendicular, CD and De form the heights of the two triangles.






We are given that
We have found in (d) that , therefore;
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